Find the best solutions to your problems with the help of IDNLearn.com. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.
Sagot :
To understand the geometric effect of the transformation matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
on vectors and areas in a 2-dimensional space, let’s analyze how it transforms some basic vectors and how it affects areas.
### 1. Effect on Unit Vectors
#### a. Unit Vector along the x-axis:
The unit vector along the x-axis is
[tex]$ \left[\begin{array}{c} 1 \\ 0 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 1 + 0 \cdot 0 \\ 0 \cdot 1 + \frac{1}{2} \cdot 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right] $[/tex]
Therefore, the unit vector along the x-axis is scaled down to
[tex]$\left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right]$[/tex].
#### b. Unit Vector along the y-axis:
The unit vector along the y-axis is
[tex]$ \left[\begin{array}{c} 0 \\ 1 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 0 \\ 1 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 0 + 0 \cdot 1 \\ 0 \cdot 0 + \frac{1}{2} \cdot 1 \end{array}\right] = \left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right] $[/tex]
Therefore, the unit vector along the y-axis is scaled down to
[tex]$\left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right]$[/tex].
### 2. Effect on Area
The area effect of a transformation matrix is determined by its determinant. The determinant of the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is:
[tex]$ \det \left( \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \right) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) - 0 \cdot 0 = \frac{1}{4} $[/tex]
This value tells us that any area in the 2D space will be scaled by a factor determined by the determinant of the matrix. In this case, areas are scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex].
### Conclusion
The geometric effect of the transformation represented by the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is to scale both the x and y components of any vector by [tex]\( \frac{1}{2} \)[/tex]. This results in vectors being halved in length along both axes. Additionally, any area in the plane is scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex], as indicated by the determinant of the transformation matrix.
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
on vectors and areas in a 2-dimensional space, let’s analyze how it transforms some basic vectors and how it affects areas.
### 1. Effect on Unit Vectors
#### a. Unit Vector along the x-axis:
The unit vector along the x-axis is
[tex]$ \left[\begin{array}{c} 1 \\ 0 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 1 \\ 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 1 + 0 \cdot 0 \\ 0 \cdot 1 + \frac{1}{2} \cdot 0 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right] $[/tex]
Therefore, the unit vector along the x-axis is scaled down to
[tex]$\left[\begin{array}{c} \frac{1}{2} \\ 0 \end{array}\right]$[/tex].
#### b. Unit Vector along the y-axis:
The unit vector along the y-axis is
[tex]$ \left[\begin{array}{c} 0 \\ 1 \end{array}\right] $[/tex]
When this unit vector is transformed by the matrix, we have:
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \left[\begin{array}{c} 0 \\ 1 \end{array}\right] = \left[\begin{array}{c} \frac{1}{2} \cdot 0 + 0 \cdot 1 \\ 0 \cdot 0 + \frac{1}{2} \cdot 1 \end{array}\right] = \left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right] $[/tex]
Therefore, the unit vector along the y-axis is scaled down to
[tex]$\left[\begin{array}{c} 0 \\ \frac{1}{2} \end{array}\right]$[/tex].
### 2. Effect on Area
The area effect of a transformation matrix is determined by its determinant. The determinant of the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is:
[tex]$ \det \left( \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] \right) = \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right) - 0 \cdot 0 = \frac{1}{4} $[/tex]
This value tells us that any area in the 2D space will be scaled by a factor determined by the determinant of the matrix. In this case, areas are scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex].
### Conclusion
The geometric effect of the transformation represented by the matrix
[tex]$ \left[\begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array}\right] $[/tex]
is to scale both the x and y components of any vector by [tex]\( \frac{1}{2} \)[/tex]. This results in vectors being halved in length along both axes. Additionally, any area in the plane is scaled down by a factor of [tex]\( \frac{1}{4} \)[/tex], as indicated by the determinant of the transformation matrix.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
